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please help and show work not just excel
purchased by a com 13-10 A $25,000 machine will be purchased by pany whose interest rate is 12%. It will cost $50 install, bu
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Answer #1

Total annual worth = - (Total capital recovery) – Annual operating cost

                              = Initial cost (A/P, i %, n) + Salvage value (A/F, i %, n) + Annual operating cost

Initial cost = $ 25,000 + $ 5,000 = $ 30,000

AW Year 1 = - $ 30,000 (A/P, 12 %, 1) + $ 16,000 (A/F, 12 %, 1) - $ 5,000

                = -$ 30,000 x 1.12 + $ 16,000 x 1 - $ 5,000

                = -$ 33,600 + $ 16,000 - $ 5,000

                = -$ 17,600 - $ 5,000 = - $ 22,600

AW Year 2 = - $ 30,000 (A/P, 12 %, 2) + $ 13,000 (A/F, 12 %, 2) - $ 5,000 + $ 3,000 (A/G, 12 %, 2)

                = -$ 30,000 x 0.5917 + $ 13,000 x 0.4717 – ($ 5,000 + $ 3,000 x 0.472)

                = -$ 17,751 + $ 6,132.10 – ($ 5,000 + $ 1,416)

                = -$ 11,618.90 - $ 6,416 = - $ 18,034.90

AW Year 3 = - $ 30,000 (A/P, 12 %, 3) + $ 11,000 (A/F, 12 %, 3) - $ 5,000 + $ 3,000 (A/G, 12 %, 3)

                = -$ 30,000 x 0.4163 + $ 11,000 x 0.2963 – ($ 5,000 + $ 3,000 x 0.925)

                = -$ 12,489 + $ 3,259.30 – ($ 5,000 + $ 2,775)

                = -$ 9,229.70 - $ 7,775 = - $ 17,004.70

AW Year 4 = - $ 30,000 (A/P, 12 %, 4) + $ 10,000 (A/F, 12 %, 4) - $ 5,000 + $ 3,000 (A/G, 12 %, 4)

                = -$ 30,000 x 0.3292 + $ 10,000 x 0.2092 – ($ 5,000 + $ 3,000 x 1.359)

                = -$ 9,876 + $ 2,301.20 – ($ 5,000 + $ 4,077)

                = -$ 7,574.80 - $ 9,077 = - $ 16,651.80

AW Year 5 = - $ 30,000 (A/P, 12 %, 5) + $ 9,500 (A/F, 12 %, 5) - $ 5,000 + $ 3,000 (A/G, 12 %, 5)

                = -$ 30,000 x 0.2774+ $ 9,500 x 0.1574 – ($ 5,000 + $ 3,000 x 1.359)

                = -$ 8,322 + $ 1,495.30 – ($ 5,000 + $ 5,325)

                = -$ 6,826.70 - $ 10,325 = - $ 17,151.70

Economic service life is 4 years as AW is least in year 4.

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